Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 16x}{x - 10} = \dfrac{3x - 90}{x - 10}$
Answer: Multiply both sides by $x - 10$ $ \dfrac{x^2 - 16x}{x - 10} (x - 10) = \dfrac{3x - 90}{x - 10} (x - 10)$ $ x^2 - 16x = 3x - 90$ Subtract $3x - 90$ from both sides: $ x^2 - 16x - (3x - 90) = 3x - 90 - (3x - 90)$ $ x^2 - 16x - 3x + 90 = 0$ $ x^2 - 19x + 90 = 0$ Factor the expression: $ (x - 9)(x - 10) = 0$ Therefore $x = 9$ or $x = 10$ At $x = 10$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 10$, it is an extraneous solution.